根据 ID 列减去行 - Pandas

Question

我有一个如下所示的数据框：

UserId    Date_watched    Days_not_watch
  1        2010-09-11         5
  1        2010-10-01         8
  1        2010-10-28         1
  2        2010-05-06         12
  2        2010-05-18         5
  3        2010-08-09         10
  3        2010-09-25         5

我想知道号码。用户给出的天数作为间隔，所以我希望每个用户的每一行都有一列，我的数据框应该如下所示：

UserId    Date_watched    Days_not_watch      Gap(2nd watch_date - 1st watch_date - days_not_watch)
  1        2010-09-11         5                0   (First gap will be 0 for all users)
  1        2010-10-01         8                15 (11th Sept+5=16th Sept; 1st Oct - 16th Sept=15days)
  1        2010-10-28         1                9
  2        2010-05-06         12               0
  2        2010-05-18         5                0 (because 6th May+12 days=18th May)
  3        2010-08-09         10               0
  3        2010-09-25         4                36
  3        2010-10-01         2                2

我在数据框的列名旁边提到了计算 Gap 的公式。

Answer 1

这是使用groupby + shift 的一种方法：

# sort by date first
df['Date_watched'] = pd.to_datetime(df['Date_watched'])
df = df.sort_values(['UserId', 'Date_watched'])

# calculate groupwise start dates, shifted
grp = df.groupby('UserId')
starts = grp['Date_watched'].shift() + \
         pd.to_timedelta(grp['Days_not_watch'].shift(), unit='d')

# calculate timedelta gaps
df['Gap'] = (df['Date_watched'] - starts).fillna(pd.Timedelta(0))

# convert to days and then integers
df['Gap'] = (df['Gap'] / pd.Timedelta('1 day')).astype(int)

print(df)

   UserId Date_watched  Days_not_watch  Gap
0       1   2010-09-11               5    0
1       1   2010-10-01               8   15
2       1   2010-10-28               1   19
3       2   2010-05-06              12    0
4       2   2010-05-18               5    0
5       3   2010-08-09              10    0
6       3   2010-09-25               5   37